Execute module one after another using flag status

Question

I want to execute 2 modules, one after another using one flag signal

What changes do I have to do in code below to the modules. I have used delay_4 and multiply.

module iir_model(yout,temp1,win,clk); input [3:0] win; output [3:0] yout; wire [3:0] yout; input clk; output [3:0] temp1; wire [3:0] temp1; delay_4 a1(temp1,win,clk); multiply_4 (yout,win,temp1); endmodule

delay_4:

module delay_4(data_out,data_in,clk); input [3:0] data_in; input clk; output [3:0] data_out; reg [3:0] data_out; reg [2:0] counter=4'b000; reg [3:0] temp; reg carry; integer i=0; always @(posedge clk) begin i=i+1; if(i==1) temp=data_in; if(i>1) begin if(counter!=4'b100) begin carry=temp[0]; temp=temp>>1'b1; temp[3]=carry; $monitor ($time," clk=%b, counter=%b ,temp=%b ,carry=%b,data_out=%b",clk, counter,temp,carry,data_out); counter=counter+1; end data_out=(temp==data_in)?temp:4'b0000; end end endmodule

multiply_4:

module multiply_4(z,x,y); input [3:0] x,y; output [7:0] z; reg [7:0] t; reg [7:0] z; integer i,j; always @ (x or y) begin z = 8'b0; for(i = 0; i < 4 ; i = i + 1) begin t = 8'b0; if(y[i]) begin t[i] = x[0]; t[i+1] = x[1]; t[i+2] = x[2]; t[i+3] = x[3]; end z = z + t; end end endmodule

In Verilog, everything in 'always' blocks executes all the time, as triggered by its '@' conditions. So multiply_4 gives you multiplied result whenever its input changes. Note that it's probably not synthesizable. — pjc50
– pjc50, Commented Mar 13, 2013 at 11:08

Martin Thompson · Accepted Answer · 2013-03-13 12:03:25Z

In Verilog, things don't execute one after the other, they execute in parallel. If you want to feed the results of one module into the inputs of another (ie a pipeline) then you just create the wires and connect them to the appropriate inputs and outputs.

I'm not even a reasonable Verilogger, so I can't see immediately what your problem is, but it might help to use named associated rather than positional association to do the wiring as its easier to see mistakes in port ordering that way.

thanks for you answer i will try using wire and if i stuck somewhere i will tell you thanks once again — Shreyas Patel
– Shreyas Patel, Commented Mar 13, 2013 at 14:07

pre_randomize · Accepted Answer · 2013-03-13 19:40:15Z

Some pointers to help you refactor code to spot the issues faster.

When posting code examples it is often nice to keep them to a minimum, and in practice less typing for you.

I would refactor the following:

module delay_4(data_out,data_in,clk); input [3:0] data_in; input clk; output [3:0] data_out; reg [3:0] data_out;

to:

module delay_4( input [3:0] data_in, input clk, output reg [3:0] data_out, );

If you do not do this then at least keep the port definitions in the same order, as the original port list, otherwise it becomes quite confusing to read, help us to help you.

I would also always have a reset that went along with the clock. I instantiate flip-flops like :

always @(posedge clk or negedge reset_n) begin if(~reset_n) begin //reset end else begin //<= next_value end end

In always @(posedge clk) block you should generally be using the non-blocking assignment <= other wise you may end up with simulator to hardware mismatches, which are quite hard to debug.

As @Martin Thompson has said named connections help aid debug, add comments showing direction for extra points.

delay_4 a1(temp1,win,clk); multiply_4 (yout,win,temp1);

would become:
NB: multiply_4 is missing an instantiation name

delay_4 a1( .data_out(temp1), //output [3:0] .data_in (win ), //input [3:0] .clk (clk ) //input ); multiply_4 multiply_4_0 ( .z (yout ), //output [7:0] .x (win ), //input [3:0] .y (temp1) //input [3:0] );

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